The Triangle - POJ 1163 水题

最新推荐文章于 2021-08-21 16:38:24 发布

提比-我有特殊的AC技巧

最新推荐文章于 2021-08-21 16:38:24 发布

阅读量670

点赞数

CC 4.0 BY-SA版权

分类专栏：水题 POJ

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/25895773

POJ 同时被 2 个专栏收录

206 篇文章

订阅专栏

水题

73 篇文章

订阅专栏

本文介绍了一个经典的算法问题——寻找从三角形顶部到底部的路径，使得路径上的数字之和最大。提供了完整的C++代码实现，并使用动态规划方法解决此问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36063		Accepted: 21565

Description

7
3   8
8   1   0
2   7   4   4
4   5   2   6   5

(Figure 1)

Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.

Input

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output

Your program is to write to standard output. The highest sum is written as an integer.

Sample Input

Sample Output

题意：从上到下一直加下来最大是多少，加的时候只能加到下面相邻的那两个。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int a[110][110];
int main()
{ int n,i,j,k,ans=0;
  scanf("%d",&n);
  for(i=1;i<=n;i++)
   for(j=1;j<=i;j++)
    scanf("%d",&a[i][j]);
  for(i=2;i<=n;i++)
   for(j=1;j<=i;j++)
    a[i][j]+=max(a[i-1][j-1],a[i-1][j]);
  for(i=1;i<=n;i++)
  if(a[n][i]>ans)
   ans=a[n][i];
  printf("%d\n",ans);
}